Sep 16, 2013

1. Computational social choice The easy-to-compute axiom Lirong Xia Sep 16, 2013

2. Topics for paper presentation is available online • On average one student one paper • group of two: two times • group of three: three times • You need to • prepare the slides • prepare the reading questions (in addition to the “what why how” questions) • prepare the talk • On the presentation day • present the paper • lead the discussion • audience: write down your answers to the reading questions • Related to your participation score

3. Last class: linear programming and computation • Linear programming • variables are positive real numbers • all constraints are linear, the objective is linear • in P • (Mixed) Integer programming • (Some) All variables are integer • NP-hard • Basic computation • Big O • Polynomial-time reduction

4. Today’s schedule • A real proof of NP-hardness (completeness) • Computational social choice: the easy-to-compute axiom • voting rules that can be computed in P • satisfies the axiom • Kemeny: a(nother) real proof of NP-hardness • IP formulation of Kemeny

5. How a reduction works? • Polynomial-time reduction: convert an instance of A to an instance of another decision problem B in polynomial-time • so that answer to A is “yes” if and only if the answer to B is “yes” • If you can do this for all instances of A, then it proves that B is HARDER than A w.r.t. polynomial-time reduction • But it does not mean that B is always harder than A P-time Instance of A Instance of B Yes Yes No No

6. NP-hard and NP-complete problems • NP-hard problems • the decision problems “harder” than any problem in NP • for any problem A in NP there exits a P-time reduction from A • NP-complete problems • the decision problems in NP that are NP-hard • the “hardest” problems in NP NP-hard NP NP-C P

7. How to prove a problem is NP-hard • How to put an elephant in a fridge • Step 1. open the door • Step 2. put the elephant in • Step 3. close the door • To prove a decision problem B is NP-hard • Step 1. find a problem A • Step 2. prove that A is NP-hard • Step 3. find a p-time reduction from A to B • To prove B is NP-complete • prove B is NP-hard • prove B is in NP (find a p-time verification for any correct answer)

8. The first known NP-complete problem • 3SAT • Input: a logical formula F in conjunction normal form (CNF) where each clause has exactly 3 literals • F = (x1∨x2∨x3)∧(¬x1∨¬x3∨x4)∧(¬x2∨x3∨¬x4) • Answer: Is F satisfiable? • 3SAT is NP-complete (Cook’s theorem)

9. Vertex cover (VC) • Vertex cover (VC): • Given a undirected graph and a natural number k. • Does there exists a set S of no more than k vertices so that every edge has an endpoint in S • Example: Does there exists a vertex cover of 4?

10. VC is NP-complete • Given F= (x1∨x2∨x3)∧(¬x1∨¬x3∨x4)∧(¬x2∨x3∨¬x4) • Does there exist a vertex cover of 4+2*3? x1 ¬x1 x2 ¬x2 x3 ¬x3 x4 ¬x4 x1 ¬x1 ¬x2 x3 x2 ¬x3 x4 x3 ¬x4

11. Notes • More details: http://cgm.cs.mcgill.ca/~athens/cs507/Projects/2001/CW/npproof.html • A yes to B must correspond to a yes to A • if yes↔nothen this proves coNP-hardness • The best source for NP-complete problems • Computers and Intractability: A Guide to the Theory of NP-Completeness • by M. R. Garey and D. S. Johnson • cited for >46k times [Google Scholar] • vs the “most cited book” The Structure of Scientific Revolutions 59K

12. The easy-to-compute axiom • A voting rule satisfies the easy-to-compute axiom if computing the winner can be done in polynomial time • P: easy to compute • NP-hard: hard to compute • assuming P≠NP

13. The winner determination problem • Given: a voting rule r • Input: a preference profile D and an alternative c • input size: n log m • Output: is c the winner of r under D?

14. Computing positional scoring rules • If following the description of r the winner can be computed in p-time, then r satisfies the easy-to-compute axiom • Positional scoring rule • For each alternative (miter) • for each vote in D(niter) • find the position of m, find the score of this position • Find the alternative with the largest score (miter) • Total time O(mn+m)=O(mn)

15. Computing the weighted majority graph • For each pair of alternatives c,d(m(m-1) iter) • let k = 0 • for each vote R • if c>d add 1 to the counter k • if d>c subtract 1 from k • the weight on the edge c→d is k

16. Kemeny’s rule • Kendall tau distance • K(R,W)= # {different pairwise comparisons} • Kemeny(D)=argminW K(D,W) =argminWΣR∈D K(R,W) • For single winner, choose the top-ranked alternative in Kemeny(D) K( b ≻c≻a,a≻b≻c ) = 2

17. Computing the Kemeny winner • For each linear order W(m! iter) • for each vote R in D(niter) • compute K(R,W) • Find W* with the smallest total distance • W*= argminWK(D,W)=argminW ΣR∈DK(R,W) • top-ranked alternative at W* is the winner • Takes exponential O(m!n) time!

18. Kemeny • Ranking R → direct acyclic complete graph G(R) • Given the WMG G(D) of the input profile D • K(D,W) =Σa→b∈G(W)(n-w(a→b))/2 = constant - Σa→b∈G(W) w(a→b)/2 • argminWK(D,W)=argmaxWΣa→b∈G(W)w(a→b) a b a>b>c>d c d

19. Kemeny is NP-hard to compute • Reduction from feedback arc set • Given a directed graph and a number k • does there exist a way to eliminate no more than k edges to obtain an acyclic graph? c c1 c1 c2 c2 c2 c2

20. Satisfiability of easy-to-compute

21. Solving Kemeny in practice • For each pair of alternatives a, b there is a binary variable xab • xab = 1 if a>b in W • xab= 0 if b>ain W • max Σa,bw(a→b)xab s.t. for all a, b, xab+xba=1 for all a, b, c, xab+xbc+xba≤2 • Do we need to worry about cycles of >4 vertices? Homework No edges in both directions No cycle of 3 vertices

22. Advanced computational techniques • Approximation • Randomization • Fixed-parameter analysis

23. Next class: combinatorial voting • In California, voters voted on 11 binary issues ( / ) • 211=2048 combinations in total • 5/11 are about budget and taxes • Prop.30 Increase sales and some income tax for education • Prop.38 Increase income tax on almost everyone for education